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3 years ago

Will give extra points

Mathematics

1 answer:

ANTONII [103]3 years ago

7 0

Answer: what is this again?

Step-by-step explanation:

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a triangle has a perimeter of 10x+2 . two sides have lengths of 5x and 2x+9 what is the length of the 3rd side

stiks02 [169]

Since all the sides add up to 10x+2, that means that 10x-2-<side1>-<side2>=<side3>. Plugging it in, we have 10x-2-5x-(2x+9)=side3, and 5x-2-(2x+9)=side3, then expanding it to 5x-2-2x-9=3x-11=side3

8 0

3 years ago

Order from least to greatest 0.5, 1/5, 0.35, 12/25, 4/5

sdas [7]

First change them all into either fractions or decimals. Decimals are probably easier though. 1/5 = 0.2; 12/25 = 0.48; 4/5 = 0.8
Order: 0.2, 0.35, 0.48, 0.5, 0.8
Then you change the decimals that you converted back into the fraction form
Answer: 1/5, 0.35, 12/25, 0.5, 4/5

5 0

3 years ago

Which statements below are true?

Alex17521 [72]

Answer:

A. Distance is always greater than or equal to the magnitude of the

displacement

Step-by-step explanation:

5 0

3 years ago

How do you solve this ?

levacccp [35]

Put the table into mahpapa.com and it will probably tell u other than that i dunno

5 0

3 years ago

About 7% of men in the United States have some form of red-green color blindness. Suppose we

Afina-wow [57]

Answer:

<h2>The probability will be $\frac{9538}{37345}$ </h2>

Step-by-step explanation:

<u>Let the total population of United States is 100.</u>

As per the given scenario, among them 7 men have the red-green color blindness.

Among the 4 selected males having at least 1 man having color blindness there could be total 4 possible cases.

CASE 1:

1 of the 4 men having blindness.

The probability will be $\frac{7C1\times93C3}{100C4} = \frac{7\times\frac{93\times92\times91}{6} }{\frac{100\times99\times98\times97}{24} } = \frac{28\times93\times92\times91}{100\times99\times98\times97}$

CASE 2:

2 of the 4 men having blindness.

The probability in this case will be $\frac{7C2\times93C2}{100C4} = \frac{7\times6\times\frac{93\times92}{2} }{\frac{100\times99\times98\times97}{24} } = \frac{56\times93\times92}{100\times99\times98\times97}$

CASE 3:

3 of the 4 men having blindness.

The required probability is $\frac{7C3\times93C1}{100C4} = \frac{7\times6\times\frac{93\times5}{6} }{\frac{100\times99\times98\times97}{24} } = \frac{35\times93\times24}{100\times99\times98\times97}$

CASE 4:

All of the 4 men that will be chosen, have the blindness.

In this case all of the men will be chosen from the 7% of the total population.

Hence, the probability is $\frac{7C4}{100C4} = \frac{7\times6\times5\times4 }{100\times99\times98\times97 } = \frac{35\times24}{100\times99\times98\times97}$

As any of the above 4 cases could be possible, in order to get the desired answer we need to add them.

hence, the answer is $\frac{9538}{37345}$

4 0

3 years ago

Will give extra points​

Will give extra points